Show that $m:\mathbb{R}^m\times\mathbb{R}^m\rightarrow\mathbb{R},(x,y)\mapsto\langle x,y\rangle$ is differentiable. Show, making use of the definition, that $m:\mathbb{R}^m\times\mathbb{R}^m\rightarrow\mathbb{R},(x,y)\mapsto\langle x,y\rangle$ is differentiable in all $(x_0,y_0)\in\mathbb{R}^m\times\mathbb{R}^m$ with $m'(x_0,y_0)(u,v)=\langle y_0,u \rangle + \langle x_0,v \rangle$. For this understand $\mathbb{R}^m\times\mathbb{R}^m$ as $\mathbb{R}^{2m}=\{(x,y):x,y\in\mathbb{R}^m\}$.

I've been looking at this close for an hour now and still don't know what to do. If I understand correctly I am supposed to show that the inner product function is differentiable. I've looked up proofs for the Leibniz rule and the Derivative of the Dot Product of Vector-Valued Functions, but I can't wrap my head around this at all. Does anybody know where I could look at the proof of specifically what is asked above? The notation $m'(x_0,y_0)(u,v)=\langle y_0,u \rangle + \langle x_0,v \rangle$ in particular confuses the hell out of me.
 A: Let $L(u,v)=\langle y_0,u\rangle+\langle x_0,v\rangle$. Then\begin{align}\frac{\bigl|m(x,y)-m(x_0,y_0)-L(x-x_0,y-y_0)\bigr|}{\|(x,y)-(x_0,y_0)\|}&=\frac{\bigl|\langle x,y\rangle-\langle x_0,y_0\rangle-\langle y_0,x-x_0\rangle-\langle x_0,y-y_0\rangle\bigr|}{\|(x-x_0,y-y_0)\|}\\&=\frac{\bigl|\langle x-x_0,y-y_0\rangle\bigr|}{\sqrt{\|x-x_0\|^2+\|y-y_0\|^2}}\\&\leqslant\frac{\|x-x_0\|.\|y-y_=\|}{\sqrt{\|x-x_0\|^2+\|y-y_0\|^2}},\end{align}by the Cauchy-Schwarz inequality. Since$$\lim_{(x,u)\to(0,0)}\frac{xy}{\sqrt{x^2+y^2}}=0,$$we have then that$$\lim_{(x,y)\to(x_0,y_0)}\frac{\bigl|m(x,y)-m(x_0,y_0)-L(x-x_0,y-y_0)\bigr|}{\|(x,y)-(x_0,y_0)\|}=0,$$which means that $m'(x_0,y_0)=L$.
A: This is a comment/hint but it wouldn't be legible in the comment box so:
It might help to relabel things and make everything explicit. Define the function:
$$f: \mathbb R^n\times \mathbb R^n \to \mathbb R, f((x_1,\dots,x_n),(y_1,\dots,y_n)) = \sum_{i=1}^nx_iy_i.$$
Note that $f$ is simply the inner product function you have defined (after fixing a basis for $\mathbb R^n$). Can you prove that this is differentiable? How about the case $n=1$? The function $f$ is just linear in each variable so things should be as simple as possible.
A: So let's start from the beginning. And using only general properties of the inner product. I mean not making assumptions on a specific inner product.
$m$ is a function from $\mathbb R^m \times \mathbb R^m$ into $\mathbb R$. So providing that $m$ is differentiable, its derivative at $(x_0,y_0)$ is a linear function from $\mathbb R^m \times \mathbb R^m$ into $\mathbb R$. This at least explains to you the notation $m^\prime(x_0,y_0)(u,v)$: at $(x_0,y_0)$ $m^\prime$ is a function of $(u,v) \in \mathbb R^m \times \mathbb R^m$.
Now... this is only the beginning. We have to prove that the derivative exists and that $m^\prime(x_0,y_0)(u,v)=\langle y_0,u \rangle + \langle x_0,v \rangle$. We have
$$
\begin{aligned}
m(x_0+u,v_0+h) &= \langle x_0+u,y_0+v \rangle\\
&=\langle x_0,y_0 \rangle+\langle x_0,v \rangle+\langle u,y_0 \rangle+\langle u,v \rangle
\end{aligned}$$
using the bilinearity of the inner product.
There is only one thing remaining to reach the end... proving that 
$$\lim\limits_{(u,v) \to (0,0)} \frac{m(x_0+u, y_0+v) - m(x_0,y_0)- m^\prime(x_0,y_0)(u,v)}{\Vert (u,v) \Vert} = 0$$
Which is just the definition of the derivative!
Rewording it with the definition of the inner product, we have to prove that
$$\lim\limits_{(u,v) \to (0,0)} \frac{\langle u,v \rangle}{\Vert (u,v) \Vert} = 0$$
Let's have Cauchy-Schwarz inequality comes into play, and the fact that for two reals $a,b$ $2\vert a \vert \vert b \vert \le \vert a \vert^2 + \vert b \vert^2$ to get
$$0 \le \vert \langle u,v \rangle\vert \le \Vert u \Vert \Vert v \Vert \le \frac{\Vert u \Vert^2 + \Vert v \Vert^2}{2\Vert (u,v) \Vert} \le \frac{\sqrt{\Vert u \Vert^2 + \Vert v \Vert^2}}{2}$$
As this last quantity tends to $0$ with $(u,v)$... we're done.
Great... Just on time for the appetizer!
